Single Machine Scheduling Problems with Financial Resource
Constraints: Some Complexity Results and Properties
Evgeny R. Gafarova , Alexander A. Lazareva,b , Frank Wernerc,∗
a
Institute of Control Sciences of the Russian Academy of Sciences, Profsoyuznaya str. 65, Moscow,
117997, Russia
b
Higher School of Economics – State University, Myasnitskaya str. 20, Moscow, 101990, Russia
c
Fakultät für Mathematik, Otto-von-Guericke-Universität Magdeburg, PSF 4120, 39016 Magdeburg,
Germany
Abstract
We consider single machine scheduling problems with a non-renewable resource. This type of
problems has not been intensively investigated in the literature so far. For several problems
of this type with standard objective functions (namely the minimization of makespan, total
tardiness, number of tardy jobs, total completion time and maximum lateness), we present
some complexity results. Particular attention is given to the problem of minimizing total
tardiness. In addition, for the so-called budget scheduling problem with minimizing the
makespan, we present some properties of feasible schedules.
Keywords: Financial scheduling, Complexity, Single machine problems, Non-renewable
resource
2000 MSC: 90B35
1. Introduction
In a resource-constrained scheduling problem, one wishes to schedule the jobs in such a
way that the given resource constraints are fulfilled and a given objective function attains
its optimal value. We consider single machine scheduling problems with a non-renewable
resource. For example, money or fuel provide natural examples of such a non-renewable
resource. Such problems with a non-renewable resource are also referred to as financial
scheduling problems. They are e.g. important for government-financed organizations. To
illustrate, it is usual for government-financed organizations that they cannot or that they do
not want to begin a project before receiving all the necessary money. For example, consider
a road-building company. Each of the projects of such a company is the construction of
one road. The company can work only on one project at the same time due to resource
∗
Corresponding author
Email addresses: [email protected] (Evgeny R. Gafarov), [email protected] (Alexander A. Lazarev),
[email protected] (Frank Werner)
Preprint submitted to Mathematical Social Sciences
April 4, 2011
constraints. However, at the beginning of a project, it must bye materials, it has to pay
subcontractors, etc. Usually, the company wants to be sure that it will receive all money
from the government in time. For this reason, all money has to be paid in advance.
The research in the area of scheduling problems with a non-renewable resource is rather
limited. In [1], some polynomially bounded algorithms are presented for scheduling problems with precedence constraints (not restricted to single machine problems). Some results
for preemptive scheduling of independent jobs on unrelated parallel machines have been presented in [11]. Toker et al. [13] have shown that the problem of minimizing the makespan
with a unit supply of a resource at each time period is polynomially solvable. Janiak et al.
[4, 5] have considered single machine problems, in which the processing times or the release
times depend on the consumption of a non-renewable resource.
The problems under consideration can be formulated as follows. We are given a set N =
{1, 2, . . . , n} of n independent jobs that must be processed on a single machine. Preemptions
of a job are not allowed. The machine can handle only one job at a time. All the jobs are
assumed to be available for processing at time 0. For each job j, j ∈ N , a processing time
pj ≥ 0 and a due date dj are given. In addition, we have one non-renewable resource G
(e.g. money, fuel, etc.) and a set of times {t0 , t1 , . . . , ty }, t0 = 0, t0 < t1 < . . . < ty , of
earnings of the resource. At each time ti , i = 0, 1, . . . , y, we receive an amount G(ti ) ≥ 0 of
the resource. For each job j ∈ N , a consumption gj ≥ 0 of the resource arises when the job
is started. Thus, we have
y
n
X
X
gj =
G(ti ).
j=1
i=0
Let Sj be the starting time of the processing of job j. A schedule S = (Sj1 , Sj2 , . . . , Sjn )
describes the order of processing the jobs: π = (j1 , j2 , . . . , jn ). Such an order is uniquely determined by a permutation (sequence) of the jobs of set N . A schedule S = (Sj1 , Sj2 , . . . , Sjn )
is feasible, if the machine processes no more than one job at a time and the resource constraints are fulfilled, i.e., for each i = 1, 2, . . . , n, we have
i
X
k=1
gjk ≤
X
G(tl ).
∀l: tl ≤Sji
Moreover, we will call the sequence π a schedule as well since one can compute S =
(Sj1 , Sj2 , . . . , Sjn ) in O(n) time applying a list scheduling algorithm to the sequence π. Then
Cjk (π) = Sjk +pjk denotes the completion time of job jk in schedule π. If Cj (π) > dj , then job
j is tardy and we have Uj = 1, otherwise Uj = 0. Moreover, let Tj (π) = max{0, Cj (π) − dj }
be the tardiness of job j in schedule π. We denote by Cmax = Cjn (π) the makespan of
schedule π and by Lj (π) = Cj (π) − dj the lateness of job j in π. In this paper, notation {π}
denotes the set of jobs contained in sequence π. Notation i ∈ π means i ∈ {π}
The single machine problem with a non-renewable resource and the minimization of the
makespan P
Cmax is denoted as 1|N R|Cmax . The
P corresponding problem of minimizing total
n
tardiness
Tj . In a similar way, we use the notations
j=1 Tj (π) is denoted as 1|N R|
2
P
P
1|N R| Cj , 1|N R| Uj and 1|N R|Lmax for the problems of minimizing total completion
time, the number of tardy jobs and maximum lateness.
The rest of this paper is organized as follows. In Section 2, we present some complexity
results P
for the single machine problems mentioned above. In Section 3, we consider problem
1|N R| Tj and prove that this problem is N P -hard even for the special case of equal
processing times. We also investigate some properties of further special cases of this problem.
The budget scheduling problem of minimizing the makespan is considered in Section 4.
2. Some Complexity Results
In Table 1, we summarize the complexity results derived in this section.
Table 1: An overview on the complexity results
Obj. Special case
func.
Cmax
P
Uj
Lmax
1)
P
Cj
P
Tj
dj = d
P
Tj
gj = g
P
Tj
pj = p
P
Tj
dj = d, gj = g
P
Tj
pj = p,
g1 ≤ g2 ≤ . . . ≤ gn ,
d1 ≤ d2 ≤ . . . ≤ dn
N P -hardness, solution algorithm
N P -hard in the strong sense (reduction from the 3-partition problem, see
Theorem 1)
N P -hard in the strong sense (Theorem 1)
N P -hard in the strong sense (Theorem 1)
N P -hard in the strong sense
(reP
duction from problem 1|rj | Cj , see
Theorem 2)
N P -hard in the strong sense (reduction from the 3-partition problem, see
Theorem 1)
P
N P -hard (problem 1|| Tj is a subcase)
N P -hard (reduction from the partition problem, see Theorem
5)
P
(in addition: 1|| Tj P∝ 1|pj =
p, G(t) = 1, t = 0, 1, . . . | Tj ) 1)
N P -hard (reduction from the partition problem, see Theorem 7)
polynomially solvable (see Remark 2),
optimal solution: π ∗ = (1, 2, . . . , n)
The notation P 1 ∝ P 2 means that problem P 1 can be polynomially reduced to P 2.
3
In this table, we refer to the partition and 3-partition problems which are as follows.
Partition problem:
A set N = {b1 , b2 , . . . , bn } of numbers is given with b1 ≥ b2 ≥ . . . ≥ bn > 0 and bi ∈ Z+ , i =
1, 2, . . . , n. Does there exist a subset N 0 ⊆ N such that
n
X
bi =
i∈N 0
1X
bi ?
2 i=1
It is known that this problem is N P -complete in the ordinary sense [3]. There exists a
pseudo-polynomial algorithm for this problem [3].
The 3-partition problem is to decide whether a given set of integers can be partitioned
into triplets that all have the same sum. More precisely:
3-Partition problem:
P
Given a set N of n = 3m positive integers b1 , b2 , . . . , bn , where ni=1 bj = mB and B4 < bj <
B
, j = 1, 2, . . . , n. Does there exist a partition of N into m subsets N1 , N2 , . . . , Nm such
2
that each subset consists exactly three numbers and the sum of the numbers in each subset
is equal, i.e.,
X
X
X
bj = B?
bj =
bj = . . . =
bj ∈N1
bj ∈N2
bj ∈Nm
The 3-partition problem is N P -complete in the strong sense [3].
Theorem 1. Problems 1|N R|Cmax , 1|N R, dj = d|
N P -hard in the strong sense.
P
Tj , 1|N R|
P
Uj and 1|N R|Lmax are
Proof. We give reductions from the 3-partition problem.
Consider first problem 1|N R|Cmax . Given the instance of the 3-partition problem, we
define an instance of problem 1|N R|Cmax as follows. There are n jobs with gj = pj =
bj , j = 1, 2, . . . , n. The times of earnings of the resource are given by {t0 , t1 , . . . , tm−1 } =
{0, B, 2B, . . . , (m − 1)B} and G(t0 ) = G(t1 ) = . . . = G(tm−1 ) = B. It is obvious that
Cmax = mB if and only if the answer for the 3-partition problem is ”YES”. In this case,
there are no idle times in an optimal
schedule.
P
P
For problems 1|N R, dj = d| Tj , 1|N R| Uj and 1|N
PR|Lmax , we define in addition to
the above data dj = d = mB for j = 1, 2, . . . , n. Then
Tj = P
0 holds if and only if the
answer for the 3-partition problem is ”YES”. Similarly, we have
Uj = 0 and Lmax = 0 if
and only if the answer for the 3-partition problem is ”YES”. Theorem 2. Problem 1|N R|
P
Cj is N P -hard in the strong sense.
Proof. It is known that the problem 1|rj |
P
Cj is N P -hard in the strong sense [9].
4
P
P
The reduction from problem
1|r
|
C
to
problem
1|N
R|
Cj can be described as
j
j
P
follows. For problem 1|rj | Cj , there are
Pgiven n jobs with r1 ≤ r2 ≤ . . . ≤ rn . We consider
the following instance of problem 1|N R| Cj . There are n jobs and the times of earnings of
the resource are given by t1 = r1 , t2 = r2 , . . . , tn = rn . Moreover, let G(ti ) = gi = 10i−1 , i =
1, 2, . . . , n. If there P
are severalPjobs l, l + 1, . . . , m − 1, m, with the same release date, then
m
j−1
we assume G(tl ) = m
. Note that we have
j=l gj =
j=l 10
k
X
gk = 1 ·
i=1
10k − 1
< 10k = gk+1 ,
10 − 1
1 ≤ k < n,
i.e., for any feasible schedule, we have Sk+1 ≥ tk+1 = rk+1 .
It is obvious that there is a one-to-one correspondence between feasible schedules forP
both
problems and the corresponding objective function values. Therefore, problem 1|N R| Cj
is N P -hard in the strong sense, too.
P
Theorem 3. Problem 1|N R, dj = d| Tj is not in AP X, where AP X is the class of
optimization problems that allow polynomial-time approximation algorithms with an approximation ratio bounded by a constant.
P
For the proof, it suffices
to
note
that
the
special
case
of
problem
1|N
R,
d
=
d|
Tj
j
P
with the optimal value
Tj = 0 is N P -hard in the strong sense.
3. Problem 1|N R|
P
Tj
P
The classical scheduling problem 1|| Tj is N P -hard in the ordinary sense
P [2, 7]. A
dynamic programming algorithm of pseudo-polynomial time complexity O(n4 pj ) has been
proposed by Lawler [6]. The state-of-the-art algorithms by Szwarc et al. [12] can solve special
instances [10]
Pof this problem for n ≤ 500 jobs. Some polynomially solvable special cases for
problem 1|| Tj have been given e.g. by Lazarev and Werner [8].
P
We start this section with the consideration of subproblem 1|N R, pj = p| Tj with equal
processing times. Then in the second
P subsection, we give a proof of N P -hardness for the
special case 1|N R,P
dj = d, gj = g| Tj . In the last subsection, we present some properties
of problem 1|N R| Tj with arbitrary processing times.
P
3.1. Special Case 1|N R, pj = p| Tj
For this special case, we can present the following trivial result:
P
Remark 1. For problem 1|N R, pj = p| Tj , there exists an optimal schedule which has
the structure π = (π1 , π2 , . . . , πy ), where the jobs in the partial schedule πi , i = 1, 2, . . . , y,
(for the definition of y, see the description of this problem in Section 1) are processed in
EDD (earliest due date) order.
In addition, we obtain the following polynomially solvable case.
5
Remark 2. For the special case of problem 1|N R, pj = p|
gn , d1 ≤ d2 ≤ . . . ≤ dn , schedule π ∗ = (1, 2, . . . , n) is optimal.
P
Tj with g1 ≤ g2 ≤ . . . ≤
The proof of polynomial solvability of this special case is trivial. For the two sequences
π = (π1 , k, l, π2 ) and π 0 = (π1 , l, k, π2 ), where l < k, we have
n
X
j=1
Tj (π) −
n
X
Tj (π 0 ) ≥ (Tk (π) + Tl (π)) − (Tk (π 0 ) + Tl (π 0 )) ≥ 0
j=1
since Cl (π 0 ) ≤ Ck (π), Ck (π 0 ) ≤ Cl (π) and dl ≤ dk .
Next, we consider
a more specific situation, namely a sub-problem denoted as 1|N R :
P
αt = 1, pj = p| Tj (see below). After proving N P -hardness
P of this special case, we consider
another special case denoted as 1|N R, G(t) = M, pj = p| Tj and derive a relation between
these
P two sub-problems. Then we give a proof of N P -hardness for problem 1|N R, pj =
p| Tj .
Now we consider the situation,
P where the times of earnings of
Pthe resource are given
by {t1 , t2 , . . . , ty } = {1, 2, . . . , gj }, t1 = 1, t2 = 2, . . . , ty =
gj , and G(ti ) = 1 for
i = 1, 2, . . . , y. This condition is P
denoted as αt = 1 [13]. Therefore, we can denote this
problem as 1|N R : αt = 1, pj = p| Tj .
Theorem 4. Problem 1|N R : αt = 1, pj = p|
P
Tj is N P -hard.
P
Proof.
We
give
the
following
reduction
from
problem
1||
Tj . Given an instance of problem
P
0
0
1|| Tj with processing times pj and duePdates dj for j = 1, 2, . . . , n, we construct an
instance of problem 1|N R : αt = 1, pj = p| Tj as follows. Let gj = p0j , pj = 0 and dj = d0j
for j = 1, 2, . . . , n. Then both problems are equivalent.
P
It P
can be noted that the special case 1|N R : αt = 1, pj = 0| P
Tj can be solved in
4
O(n
gj ) time by Lawler’s algorithm [6] since we obtain a problem 1|| Tj with processing
times gj . As we have already noted,
P later in Theorem 6, we will present another N P -hardness
proof for problem 1|N R : pP
=
p|
Tj . The reason for this is as follows. Let I be an instance
j
of problem 1|N R : pj = p| Tj and x be a string of the form
”p, d1 , d2 , . . . , dn , g1 , g2 , . . . , gn , t0 , t1 , . . . , ty , G(t0 ), G(t1 ), . . . , G(ty )”
encoding the instance I under a reasonable encoding scheme e. According to the definition
in [3], we have LEN GT H[I] = n + y. In fact, the string x consists
P of 2n + 2y + 3 numbers.
However, if we consider the
problem
1|N
R
:
α
=
1,
p
=
0|
Tj as a special case of the
t
j
P
problem 1|N R
P : pj = p| Tj and use the same encoding scheme, then LEN GT H[I] =
n + y = n + gj , i.e., the length of the P
input is pseudo-polynomial.
P Since, as mentioned
above, the problem 1|N R : αt = 1, pj = 0| Tj can be solved in O(n gj ) time by Lawler’s
algorithm, the complexity
P of this algorithm would polynomially depend on the input length
LEN GT H[I] = n + gj . For this reason, we consider sub-problem 1|N R : αt = 1, pj =
6
P
p| Tj as a separate problem and use the encoding scheme e0 , in which we present an
instance as a string
”p, d1 , d2 , . . . , dn , g1 , g2 . . . , gn ”,
i.e., LEN GT H[I] = n.
P
We consider the sub-case of problem 1|N R, pj = p| Tj , where the times of earnings
of the resource are given
P by t1 = M, t2 = 2M, . . . , tn = nM and G(ti ) = M for all i =
g
1, 2, . . . , n, P
where M = n j such that M ∈ Z+ . We denote this special case by 1|N R, G(t) =
M, pj = p| Tj .
P
Two
instances
of
problems
1|N
R
:
α
=
1,
p
=
p|
Tj and 1|N R, G(t) = M, pj =
t
j
P
p| Tj are called corresponding, if all parameters dj , pj , gj , j = 1, 2, . . . , n, for the two
instances are the same.
P
Now we investigate a relation between the values of the objective function
Tj for two
corresponding instances. First, we show that these two instances can have different optimal
sequences (see Remark 3). Then
of dj = 0 for j = 1, 2, . . . , n
P we consider the special case P
such that objective function Tj turns into the special case of Cj , and we investigate the
difference between the function values of the same sequence for the two problems mentioned
above (see Remarks 4 and 5).
Remark
3. There exist two corresponding
instances of problems 1|N R : αt = 1, pj =
P
P
p| Tj and 1|N R, G(t) = M, pj = p| Tj which have different optimal schedules.
We consider an instance with n =P2 jobs and p1 =
2 = 5, d1 = 7, d2 = 6.
Pp2 = 11, g1 = 1, gP
For problem 1|N R : αt = 1, pj = p| Tj , we have
Tj (π ) = 0 and
Tj (π 2 ) = 1, where
2
π 1P
= (2, 1) and πP
= (1, 2). On the P
other hand, for the problem 1|N R, G(t) = M, pj =
p| Tj , we have
Tj (π 1 ) = 2 and
Tj (π 2 ) = 1. Thus, the above two instances have
different optimal schedules.
Now, let dj P
= 0 for j = 1, 2, . . . , n. For two corresponding
instances of problems 1|N R :
P
αt = 1, pj = 1| Cj and 1|N R, G(t) = M, pj = 1| Cj , let Cj (π) be the completion time of
job j according to the job sequence π for the first problem and Cj0 (π) be the completion time
of the same job according to π for the second problem. Then we can prove the following
remark.
Remark 4. For two corresponding
instances of problems 1|N R : αt = 1, pj = 1|
P
1|N R, G(t) = M, pj = 1| Cj , we have
Pn
C 0 (π)
Pnj=1 j
< 2.
j=1 Cj (π)
P
Cj and
(1)
Proof. First, we show that inequality
Cj0 (π) − Cj (π) ≤ M − 1
holds for each job j, j = 1, 2, . . . , n. We denote Sj0 (π) = Cj0 (π) − pj = Cj0 (π) − 1.
7
(2)
Let π = (1, 2, . . . , i, i + 1, . . . , j, . . . , n). Then there exists a natural number k such that
Si−1 (π) < (k − 1)M ≤ Su (π) < kM
for u = i, i + 1, . . . , j. In addition, it is obvious that
Su0 (π) < (k + 1)M
for u = i, i + 1, . . . , j. If the jobs i, i P
+ 1, . . . , j are processed from time kM in schedule π
for problem 1|N R, G(t) = M, pj = 1| Cj , then Si (π) > (k − 1)M and
X
Sj0 (π) − kM =
pl = j − i < Sj (π) − (k − 1)M.
l∈{i,i+1,...,j−1}
Thus, inequality (2) holds.
Moreover, we have Cmax (π) ≥ nM + 1 since t = nM is the last time of earnings of the
0
resource, i.e., nM is a lower bound for Cmax and Cmax
. Hence, we get
n
X
Cj0 (π) −
n
X
j=1
Cj (π) ≤ n(M − 1) < Cmax (π) <
j=1
n
X
Cj (π).
j=1
Therefore, inequality (1) holds.
Remark 5. There existsPan instance of the problems 1|N R : αt = 1, pj = 1|
1|N R, G(t) = M, pj = 1| Cj for which we have
Pn
Cj0 (π)
1
Pnj=1
≈2− .
n
j=1 Cj (π)
P
Cj and
Proof. Let us consider the following instance. There areP
given n jobs with gj = 1, j =
3
1, 2, . . . , n − 1 and gn = nM − (n − 1), where M = n
j∈N \{n} gj . For schedule π =
(1, 2, . . . , n), using inequality (2) in the proof of Remark 4, we obtain
n−1
X
j=1
Cj0 (π) −
n−1
X
Cj (π) = (n − 1)(M − 1),
j=1
Moreover, we have
Cn = Cmax = Cn0 = nM + 1.
This yields
n
X
Cj (π) = 2 + 3 + . . . + n + nM + 1
j=1
=
n(n + 1)
− 1 + nM + 1.
2
8
Let n be fixed. Then we obtain
Pn
0
j=1 Cj (π)
P
=
lim
n
M →∞
j=1 Cj (π)
lim
(n − 1)(M − 1) +
n(n+1)
2
M →∞
n(n+1)
2
+ nM
+ nM
2n − 1
n
1
= 2− .
n
=
P
Now we consider the sub-case of the problem 1|N R, pj = p| Tj , where the number of
times of earnings of the resource given by t0 , t1 , . . . , ty is less than or equal to n, i.e., y ≤ n
(we remind the discussion after the proof of Theorem 4).
P
Theorem 5. The special case 1|N R, pj = p| Tj , where the number of times of earnings
of the resource given by t0 , t1 , . . . , ty is less than or equal to n, is N P -hard.
Proof. The reduction is done from the partition problem. We consider an instance of
the scheduling problem with n jobs and pj = 0, gj = bj for all j = 1, 2, . . . , P
n. Two times
of earnings of the resource t0 = 0 and t1 = M are given, where M = (n nj=1 bj )2 and
P
G(t0 ) = G(t1 ) = 12 nj=1 bj . In addition, the due dates dj = M − gj for j = 1, 2, . . . , n are
given.
Then all optimal schedules have the form π = (E, F ) with the following property. For
all jobs j ∈ E, we have
n
Cj (π) = 0,
Tj (π) = 0,
X
gj ≤
j∈E
1X
bj
2 j=1
and for all jobs j ∈ F , we have
Cj (π) = M,
Tj (π) = M − dj = gj .
Then the instance of the partition problem
has an answer ”YES” if and only if in an optimal
P
schedule π of problem 1|N R, pj = p| Tj , we have
n
X
n
1X
Tj (π) =
bj .
2 j=1
j=1
P
As an immediate corollary of Theorem 5 we obtain that problem 1|N R, pj = p| Tj is
N P -hard which is the result presented in Table 1.
9
P
3.2. Special Case 1|N R, dj = d, gj = g| Tj
P
In this subsection, a proof of N P -hardness for the special case 1|N R, dj = d, gj = g| Tj
is presented.
P
We give the following reduction from the partition problem. Denote M = (n nj=1 bj )n .
Let us consider the following instance with the set of jobs N = {1, 2, . . . , 2n + 1}:

p2n+1 = 1,




p2i = M n−i+1 ,
i = 1, 2, . . . , n,




p
=
p
+
b
,
i
= 1, 2, . . . , n,

i P
 2i−1Pn 2i

1

d = i=1 p2i + 2 bj ,




g = 1,




G(t0 ) = n,
 t0 = 0,
t1 = d, P
G(t1 ) = 1,
(3)


t
=
t
+
b
+
1,
G(t
)
=
1,

2
1
j
2



t
=
t
+
p
+
b
,
G(t

3
2
2n
n
3 ) = 1,



...
...




t
=
t
+
p
+
b
,
G(t

i
i−1
n−i+3
i ) = 1,
2(n−i+3)



...
...



tn+1 = tn + p4 + b2 ,
G(tn+1 ) = 1.
It is obvious that there are at least n + 1 tardy jobs in any feasible schedule. We define
a canonical schedule as a schedule of the form
(V1,1 , V2,1 , . . . , Vi,1 , . . . , Vn,1 , 2n + 1, Vn,2 , . . . , Vi,2 , . . . , V2,2 , V1,2 ),
where
{Vi,1 , Vi,2 } = {2i − 1, 2i},
i = 1, 2, . . . , n.
Moreover, let π = (E, F ) and for the two partial schedules E and F , we have |{E}| = n and
|{F }| = n + 1. Note that in any canonical schedule, all jobs in sub-sequence F are tardy,
the last job in sub-sequence E can be tardy or on-time while all other jobs in sub-sequence
E are on-time.
Theorem 6. For instance (3), there exists an optimal schedule which is canonical.
P
Proof. Without loss of generality, assume that n > 2 and
bj > 2.
(1) Let us first prove that job V1,2 (i.e., one of the jobs 1 or 2) is the last job in an optimal
schedule. Assume
P that there exists an optimal schedule π = (π1 , V1,1 , π2 , V1,2 , π3 , j).
Denote P1 = i∈N \{V1,1 ,V1,2 } pi . It is obvious that CV1,2 (π) > 2M n > tn+1 . Moreover,
2M n − tn+1 > 2P1 .
Then, for schedule π 0 = (π1 , V1,1 , π2 , π3 , j, V1,2 ), we have
2n+1
X
j=1
Tj (π) −
2n+1
X
Tj (π 0 ) ≥ (TV1,2 (π) − TV1,2 (π 0 )) + (Tj (π) − Tj (π 0 )) >
j=1
> −P1 + (CV1,2 (π) − tn+1 ) >
> −P1 + 2M n − tn+1 > P1 > 0.
10
Thus, schedule π is not optimal, and job V1,2 is the last job in an optimal schedule.
(2) We prove that in an optimal schedule π = (E, F ), job V1,1 belongs to sub-sequence E
with |{E}| = n. Assume that π = (π1 , π2 , V1,1 , π3 , V1,2 ), where π1 = E. It is obvious
that d − P1 > 0 and CV1,2 (π) − tn+1 > 2M n + d − tn+1 > 2M n − P1 .
S
S
Let us consider schedule π 0 = (V1,1 , π1 , π2 , π3 , V1,2 ). For each job i ∈ {π1 } {π2 } {π3 },
we have Ti (π 0 ) − Ti (π) < P1 . Moreover, TV1,2 (π) − TV1,2 (π 0 ) > M n − P1 . Then
2n+1
X
Tj (π) −
j=1
2n+1
X
Tj (π 0 ) > M n − P1 − (2n − 1)P1 > 0,
j=1
since M n > 21 M ·P1 > 2nP1 . Thus, in an optimal schedule π = (E, F ), job V1,1 belongs
to sub-sequence E with |{E}| = n.
(3) Let us show that there exists an optimal schedule π = (E, F ), where job V1,1 is the
first job.
(3.1) First, we prove that in an optimal schedule, only one of the jobs V2,1 and V2,2 can
be in sub-sequence E. In contradiction, without loss
assume that
S of generality,
S
π = (V2,1 , V2,2 , π1 , V1,1 , j, π2 , V1,2 ), where {V2,1 , V2,2 } {π1 } {V1,1 } = {E} and j
is the first job in sub-sequence F . From the previous item, it is obvious that there
exists an optimal schedule, where V1,1 is sequenced
on position n since job V1,1
P
belongs to E with |{E}| = n. Denote P = i∈{V2,1 ,V2,2 } S{π1 } S{V1,1 } .
If P − pV2,2 ≤ d, then schedule π 0 = (V1,1 , V2,1 , π1 , V2,2 , j, π2 , V1,2 ) is also optimal,
and job V1,1 is the first job.
Otherwise, if P −pV2,2 > d, then Tj (π) > pV2,2 and TV1,1 (π) > pV2,2 . Let us consider
schedule π 0 = (V2,1 , π1 , V1,1 , j, V2,2 , π2 , V1,2 ). It is easy to show that CV2,2 (π 0 ) =
Cj (π) < d + 2pV2,2 . Thus, we have
2n+1
X
Tj (π) −
j=1
2n+1
X
Tj (π 0 ) > 0.
j=1
Now it is easy to show that for schedule π 00 = (V1,1 , π1 , V2,1 , j, V2,2 , π2 , V1,2 ), we
have
2n+1
2n+1
X
X
00
Tj (π ) =
Tj (π 0 )
j=1
since
P
i∈{V2,1 }
S
{π1 }
S
{V1,1 }
j=1
pi − d < pV2,1 , and so job V1,1 is the first job.
(3.2) If {V2,1 , V2,2 } ∈
/ E, then all jobs from sub-sequence E are not tardy in any
schedule of the type π = (E, F ), where V1,2 ∈ F . Thus, job V1,1 is the first job,
too.
11
(4) Now we consider only optimal schedules of the type π = (V1,1 , π1 , π2 , V1,2 ). We use the
same three steps (1) to (3) to prove that there exists an optimal schedule of the type
π = (V1,1 , V2,1 , π10 , π20 , V2,2 , V1,2 ).
(5) Then, by induction, we assume that there exists an optimal schedule of the type
π = (V1,1 , V2,1 , . . . , Vi−1,1 , π1 , π2 , Vi−1,2 , . . . , V2,2 , V1,2 ), i ≥ 3, and we prove according to
steps (1) - (3) that there exists an optimal schedule
π = (V1,1 , V2,1 , . . . , Vi−1,1 , Vi,1 , π10 , π20 , Vi,2 , Vi−1,2 , . . . , V2,2 , V1,2 ).
Thus, the theorem has been proven.
We note that in a canonical schedule, there are either n + 1 or n + 2 tardy jobs (job Vn,1
can be tardy or on-time). Moreover, as we prove in the following theorem, in an optimal
canonical schedule, there are only n + 1 tardy jobs and thus, all jobs in sub-sequence E are
on-time.
Theorem 7. The instance of the partition problem has an answer ”YES” if and only if in
an optimal canonical schedule, equality
X
pj = d
j∈E
holds.
Proof. For a canonical schedule π, we have
2n+1
X
Tj (π) = TVn,1 + T2n+1 +
j=1
= TVn,1 + T2n+1 +
n
X
i=1
n
X
TVi,2
(tn−i+2 + p2i + φ(i)bi ),
i=1
where
φ(i) =
In addition, inequalities 0 ≤ TVn,1
we want to minimize the value
1, if Vi,2 = V2i−1 ,
0, if Vi,2 = V2i .
P
≤ 12 bi and 1 ≤ TV2n+1 ≤
TVn,1 + T2n+1 +
n
X
1
2
φ(i)bi .
i=1
We attain the minimal value
TVn,1 + T2n+1 +
n
X
i=1
n
1X
bi
φ(i)bi = 0 + 1 +
2 i=1
12
P
bi + 1 hold. In fact,
if and only if CVn,1 = d, i.e.,
X
pj = d.
j∈E
P
Thus, the special case 1|N R, dj = d, gj = g| Tj is N P -hard.
P
3.3. Properties of the problem 1|N R| Tj
We complete this section with some comments about the problem
P of minimizing total tardiness with arbitrary processing times. For this problem 1|N R| Tj , Lawler’s proposition
[6]
min(Cj∗ , dj ) ≤ d0j ≤ max(Cj∗ , dj )
holds. However, the well-known elimination rule by Emmons [12], i.e.,
pj > pi , dj ≥ di ⇒ (i → j)
does not hold. Here, the notation (i → j) means that there exists an optimal schedule in
which the processing of job i precedes the processing of job j.
Let us consider an instance with n = 3 jobs and p1 = p > 1, p2 = p3 = 1, d1 = d2 =
d3 = p + 2, g1 = 3, g2 = g3 = 2, and G(0) = 3, G(p) = 4. For all optimal schedules
of this instance, we have (j → i), pj > pi , dj ≥ di . Therefore,
Pwe cannot use the exact
pseudo-polynomial
algorithm by Lawler [6] for the problem 1|| Tj to solve the problem
P
1|N R| Tj .
4. Budget Scheduling Problems with Makespan Minimization
A budget scheduling problem is a financial scheduling problem described in this paper,
where instead of the values gj , values gj− ≥ 0 and gj+ ≥ 0 are given. The value gj− has the
same meaning as gj in the financial scheduling problem. However, at the completion time
of job j, one has additional earnings gj+ of the resource.
If we have gj− ≥ gj+ for all j = 1, 2, . . . ,P
n, then the new instance
with gj = gj− − gj+ is not
P
equivalent to the original one. Let G = nj=1 (gj− − gj+ ). If ∀t G(t) < G + max gj− , then
not all sequences (schedules)
π are feasible. For example, let n = 2 and g1− = 100, g1+ =
P
3, g2− = 3, g2+ = 2 and ∀t G(t) = 100. Then schedule (1, 2) is feasible but schedule (2, 1) is
not feasible.
We denote this problem as 1|N R, gj− , gj+ |Cmax . It is obvious that this problem is N P hard in the strong sense (since the financial scheduling problem is a special case of the
budget scheduling problem).
Remark 6. If there exists a feasible schedule for an instance of problem 1|N R, gj− , gj+ , gj− >
gj+ |Cmax , then schedule π = (1, 2, . . . , n), g1+ ≥ g2+ ≥ . . . ≥ gn+ is feasible, too.
13
If inequality gj− > gj+ does not hold for all j = 1, 2, . . . , n, then we can use the following
list scheduling algorithm for constructing a feasible schedule.
Algorithm A. First, all jobs j ∈ N with gj+ − gj− ≥ 0 are scheduled. In particular, schedule
among these jobs the job with the minimal value gj− , if there is more than one job with this
property, select the job with the largest value gj+ − gj− . If all jobs j with gj+ − gj− ≥ 0 have
been sequenced, schedule the remaining jobs according to non-increasing values gi+ .
Moreover, we can give the following remark concerning the approximability of the problem under consideration.
Remark 7. Problem 1|N R, gj− , gj+ |Cmax is in AP X.
Proof. Denote the set of times of earnings of the resource by {t1 , t2 , . . . , tl−1 , tl , . . . , tk }.
Assume that for the times {t1 , t2 , . . . , tl−1 , tl } of earnings of the resource, there already exists
a feasible schedule, but for the set {t1 , t2 , . . . , tl−1 }, there is no feasible schedule.
PnThen tl
∗
. In addition,
is a lower bound for the optimal objective function value Cmax
j=1 pj is
∗
also a lower bound for Cmax . We can construct a feasible job sequence (schedule) π using
Algorithm A, where all jobs are processed from time tl without idle times. Thus, we obtain
Cmax (π) = tl +
n
X
∗
pj ≤ 2Cmax
.
j=1
5. Concluding Remarks
In this paper, we considered single machine financial and budget scheduling problems.
For the first class of problems with a non-renewable resource, we presented various complexity results. Then we considered several special cases of the problem of minimizing total
tardiness. In addition, we presented some properties of feasible solutions for the budget
scheduling problem of minimizing the makespan.
For future research, it is interesting to derive several elimination rules and properties
of optimal solutions for the above mentioned single machine problems since most of the
problems considered are N P -hard in the strong sense and thus the construction of pseudopolynomial algorithms seems to be impossible. Another direction of future research is the
consideration of shop and parallel machine problems with a non-renewable resource.
Acknowledgements
Partially supported by DAAD (Deutscher Akademischer Austauschdienst): A/08/80442/Ref.
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